If the sum of first 2n terms of the A.P. 2, 5, 8, …, is equal to the sum of first n terms of the A.P. 57, 59, 61,…., then n equals

Sum to $2 n$ terms of 2, 5, 8,,,… is

$\frac{2 n}{2} [2 (2) + (2 n - 1) (3)] = n (6 n + 1)$

and sum to n terms of 57, 59, 61,…., is

$\frac{n}{2} [2 (57) + (n - 1) (2)] = n (56 + n)$

According to given condition

$n (6 n + 1) = n (56 + n) \Rightarrow 6 n + 1 = 56 + n \Rightarrow n = 11$

If the sum of first $2 n$ terms of the A.P. 2, 5, 8, …, is equal to the sum of first $n$ terms of the A.P. 57, 59, 61,…., then $n$ equals